Show that abs(A+B)(A-B)<=abs(A)sqr+abs(B)sqr
Let's start by expanding the left side of the inequality:
(abs(A+B)(A-B)) = (abs(A) + abs(B))(A-B)
Using the distributive property, we can expand this expression:
= abs(A)(A-B) + abs(B)(A-B)
Using the distributive property again, we have:
= A*abs(A) - B*abs(A) + A*abs(B) - B*abs(B)
Now, let's expand the right side of the inequality:
(abs(A))^2 + (abs(B))^2 = A*abs(A) + B*abs(B)
Now, let's compare the two sides of the inequality:
A*abs(A) - B*abs(A) + A*abs(B) - B*abs(B) ≤ A*abs(A) + B*abs(B)
Rearranging the terms, we have:
-A*abs(A) - B*abs(B) ≤ 0
Multiplying both sides by -1, we get:
A*abs(A) + B*abs(B) ≥ 0
Since this inequality holds true for all values of A and B, we can conclude that:
abs(A+B)(A-B) ≤ (abs(A))^2 + (abs(B))^2
Therefore, abs(A+B)(A-B) is less than or equal to abs(A)^2 + abs(B)^2, which proves the inequality.
To show that |A + B||A - B| ≤ |A|^2 + |B|^2, we can start by expanding the left side of the inequality.
First, let's consider the term |A + B|. This represents the absolute value of the sum of A and B.
|A + B| = |(A + B)| = sqrt((A + B)^2) by definition of absolute value.
Expanding the squared term gives us:
|A + B| = sqrt(A^2 + 2AB + B^2)
Next, let's consider the term |A - B|. This represents the absolute value of the difference between A and B.
|A - B| = |(A - B)| = sqrt((A - B)^2) by definition of absolute value.
Expanding the squared term gives us:
|A - B| = sqrt(A^2 - 2AB + B^2)
Now, let's multiply the two terms together:
|A + B||A - B| = sqrt(A^2 + 2AB + B^2) * sqrt(A^2 - 2AB + B^2)
Using the property of square roots that sqrt(a) * sqrt(b) = sqrt(a * b), we can simplify the expression:
|A + B||A - B| = sqrt((A^2 + 2AB + B^2)(A^2 - 2AB + B^2))
Expanding the product gives us:
|A + B||A - B| = sqrt(A^4 - 4A^2B^2 + 4AB^2 + 4A^2B^2 - 4B^4)
Simplifying further:
|A + B||A - B| = sqrt(A^4 - 4B^4)
Since the square root of a square is always positive, we can remove the square root, and we have:
|A + B||A - B| = A^2 - 4B^2
Now, let's consider the right side of the inequality: |A|^2 + |B|^2. This represents the sum of the squares of A and B.
|A|^2 + |B|^2 = A^2 + B^2
Finally, let's compare the left and right sides of the inequality:
We have |A + B||A - B| = A^2 - 4B^2 and |A|^2 + |B|^2 = A^2 + B^2
We need to show that |A + B||A - B| ≤ |A|^2 + |B|^2, which means that A^2 - 4B^2 ≤ A^2 + B^2
Simplifying the inequality gives us:
-4B^2 ≤ B^2
This inequality is always true since -4B^2 is always less than or equal to B^2.
Therefore, we have shown that |A + B||A - B| ≤ |A|^2 + |B|^2.